Question

0.924 moles of A (g) is placed in a 1.00 L container at 700 degrees Celsiusand a reaction occured. 38.8% of the gas A dissociated when equilibrium was established. 3 A (g) ⇌ 5 B (g) + 2 C (g). What is the value of the equilibrium constant, K, at this temperature? (Answer: K = 0.0241)

**PLEASE SHOW WORK. THANKS**

Answer 1

                [A] = moles / volume of container in Litres = 0.924 mol / 1L = 0.924 M           

                                  3 A (g)   ⇌      5 B (g)     +   2 C (g)

Initial                           0.924 M               0                0

At equilibrium            0.924 - 3x            5x               2x

Given that 38.8% of the gas A dissociated .

      3x = 38.8% of 0.924 M = 0.358 M

        x = 0.119 M

Therefore, Equilibrium concentrations are

[A] = 0.924 - 3x = 0.924 - 0.358 = 0.566 M

[B] = 5 x = 5 (0.119 ) = 0.595 M

[C] = 2x = 2 x 0.119 = 0.238 M

Equilibrium constant

   K = [B]⁵ [C]² / [A]³

      = [0.595]⁵ [0.238]² / [0.566]³

      = 0.0241

K = 0.0241